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\title{微分方程数值解\ 第11次作业}
\author{李之琪 22235056}

\begin{document}
\maketitle
\section*{}
对leap-frog策略，有
\begin{eqnarray}
  \begin{aligned}
    v^{n+1} = v^{n-1} + 2k(D_0v^n-v^n)
    = v^{n-1} + 2k(\frac{v^{n+1}-v^{n-1}}{2h}-v^n),
  \end{aligned}
\end{eqnarray}
从而
\begin{eqnarray}
  \begin{aligned}
    v^{n+1} 
    = v^{n-1} - \frac{2k}{1-\mu}v^n,
  \end{aligned}
\end{eqnarray}
这里$\mu = \dfrac{k}{h}$，根据CFL条件，可以认为$\mu$是一个常数。于是我
们将上式写为
\begin{eqnarray}
  \begin{aligned}
    \begin{bmatrix}
 v^{n+1}\\
 v^n
\end{bmatrix} =
\begin{bmatrix}
- \frac{2k}{1-\mu}& 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
v^n \\
v^{n-1}
\end{bmatrix}=:
Q\begin{bmatrix}
v^n \\
v^{n-1}
\end{bmatrix}.
  \end{aligned}
\end{eqnarray}
根据$S_h(t,t_0)$的定义，为了说明$\left\|S_h(t,t_0)\right\|_h \approx
e^{t}$，只需要说明$\left\| Q\right\|_h \approx e^{k}$。注意到
\begin{eqnarray}
  \begin{aligned}
    \left\| Q\right\|_h &\approx \left\| Q\right\|_2\\
    &= \frac{k}{1-\mu}+\sqrt{\frac{k^2}{(1-\mu)^2}+1}\\
    &=
    \frac{k}{1-\mu}+\left(1+\frac{1}{2}\left(\frac{k}{(1-\mu)}\right)^2
      + O\left(\frac{k}{(1-\mu)}\right)^4 \right)\\
    &\approx e^{\frac{k}{1-\mu}} \approx e^k.
  \end{aligned}
\end{eqnarray}
同理，对
\begin{eqnarray}
  \begin{aligned}
   (I+k) v^{n+1} =  2kD_0v^n + (I-k) v^{n-1},
  \end{aligned}
\end{eqnarray}
整理得
\begin{eqnarray}
  \begin{aligned}
   v^{n+1} =  \frac{1-k-\mu}{1+k-\mu}v^{n-1}.
  \end{aligned}
\end{eqnarray}
故只需要说明$\left|\dfrac{1-k-\mu}{1+k-\mu} \right| \approx e^{-2k}$。
注意到
\begin{eqnarray}
  \begin{aligned}
    \frac{1-k-\mu}{1+k-\mu} =
    \frac{1-\frac{k}{1-\mu}}{1+\frac{k}{1-\mu}} \approx
    e^{\frac{-2k}{1-\mu}} \approx e^{-2k}
  \end{aligned}
\end{eqnarray}
完成证明。
\end{document}

